For $2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$ the difference in change in enthalpy and change in internal energy at $25^{\circ} C$ (kJ) is:

-7.43

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+3.72

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-3.72

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+7.43

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Solution

The correct option is D +7.43
For $2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
$Δ n = 12 - 15 = - 3$
$Δ H = Δ E + Δ n R T$
$Δ H - Δ E = Δ n R T = - 3 \times 8.314 \times 10^{- 3} \times 298 = - 7.432 k J$

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